en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=ty^{1/2},y(0)=0

Solution

y^{^{'}} = t y^{\frac{1}{2}}, y (0) = 0

Solution

y = \frac{t ^{4}}{16}

Solution steps

y^{'} (t) = t y^{\frac{1}{2}}

Solve separable ODE:

y = \frac{t ^{4}}{16}

y = \frac{t ^{4}}{16}

Graph

Popular Examples

(dy)/y =x^{13}dx

\frac{d y}{y} = x^{13} d x

(dy)/(dx)y=sqrt(9x+7)

\frac{d y}{d x} y = 9 x + 7

y^'=(e^{4x+y}-sec(7x)e^y)

y^{^{'}} = (e^{4 x + y} - sec (7 x) e^{y})

3 y^{^{'}} + 4 y = 2

y^'+((1+y^2)/y)sin(x)=0,y(0)=2

y^{^{'}} + (\frac{1 + y ^{2}}{y}) sin (x) = 0, y (0) = 2

Frequently Asked Questions (FAQ)

What is the solution for y^'=ty^{1/2},y(0)=0 ?
The solution for y^'=ty^{1/2},y(0)=0 is y=(t^4)/(16)